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Hyperbola Calculator: Foci, Vertices and Asymptotes

Q: How do I enter 16x² − 9y² = 144?

Divide by 144: x²/9 − y²/16 = 1. Then a = 3, b = 4, center (0, 0), opening left and right.

From the center, a, b and orientation of a hyperbola get its standard equation, vertices, foci, asymptote equations, eccentricity, latus rectum and directrices.

Center x (h)

Center y (k)

a (under the positive term)

b (under the negative term)

Opens

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How it works

(x − h)²/a² − (y − k)²/b² = 1 · c² = a² + b² · vertices (h ± a, k) · foci (h ± c, k) · asymptotes y − k = ±(b/a)(x − h) · e = c/a > 1 (swap x and y for a vertical hyperbola)

A hyperbola is the set of points whose distances to two foci differ by the constant 2a. In standard form the positive squared term decides the direction: an x term first means the branches open left and right. The vertices are a units from the center, the foci c units away with c² = a² + b² (a plus sign, unlike the ellipse), and the branches approach two asymptotes through the center with slopes ±b/a (±a/b for a vertical hyperbola). The eccentricity c/a is always greater than 1.

Worked example

For (x − 1)²/9 − (y − 2)²/16 = 1: center (1, 2), a = 3, b = 4, so c = √(9 + 16) = 5. The vertices are (−2, 2); (4, 2) and the foci (−4, 2); (6, 2). The asymptotes y − 2 = ±(4/3)(x − 1) are y = (4/3)x + 2/3 and y = −(4/3)x + 10/3. The eccentricity is 5/3 ≈ 1.666667 and the latus rectum 2 × 16/3 ≈ 10.6667.

Sources: Hyperbola — Wikipedia · OpenStax Precalculus 2e — The hyperbola

Frequently asked questions

How do I tell which way a hyperbola opens?

Look at the positive term. If (x − h)² comes first with a plus sign, it opens left and right; if (y − k)² is positive, it opens up and down. Unlike an ellipse, a is not necessarily larger than b.

How do I find the asymptotes?

Replace the 1 on the right with 0 and solve: (x − h)²/a² = (y − k)²/b² gives y − k = ±(b/a)(x − h). For a vertical hyperbola the slopes are ±a/b.

Why is c² = a² + b² here but a² − b² for an ellipse?

For a hyperbola the foci lie outside the vertices, so c > a, and the defining difference of distances leads to c² = a² + b². For an ellipse the foci lie inside, so c < a.

How do I enter 16x² − 9y² = 144?

Divide by 144: x²/9 − y²/16 = 1. Then a = 3, b = 4, center (0, 0), opening left and right.

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This calculator is for educational purposes. Double-check important results before acting on them.